The Gateway Arch in St. Louis was designed by Eero Saarinen and was constructed using the equation y = 211.49 − 20.96 cosh(0.03291765x) for the central curve of the arch, where x and y are in meters and  |x| ≤ 91.20.   a-What is the height (in m) of the arch at its center? (Round your answer to two decimal places.)   b-At what points is the height 50 m? (Round your answers to two decimal places.) smaller x-value (x, y)= larger x-value (x, y)=   c-What is the slope of the arch at the points in part b? (Round your answers to one decimal place.)

Step:-1

Given equation is $y=211.49 - 20.96 \cos h \left(0.03291765 x\right)$ and x, y are in meters and $\left|x\right|\le 91.20$

Sketching the graph using given equation, we get

This is parabola opens downward.

Part (a):-

The height of arch at its center is

$$\begin{gathered} y=211.49 - 20.96 \cos h \left(0.03291765 x\right) \\ \text{put x=0 for height} \\ y=211.49 - 20.96 \cos h \left(0.03291765 \times 0\right) \\ y=211.49 - 20.96 \cos h \left(0\right) \\ \cos h \left(0\right)=1 \text{then} \\ y=211.49 - 20.96\times 1=190.53 \\ y=190.53 \end{gathered}$$

So, height is 190.63 meter.

or 

Using graph, we can see the height of arch is 190.53 meter.

Answer:-

height = 190.63 meter.

Step:-2

Part (b):-

Given that  height is 50 m.

Now, we have to find the value of x for which arch attains height 50 m.

Given equation is 

$$\begin{gathered} y=211.49 - 20.96 \cos h \left(0.03291765 x\right) \\ \text{put y=50 for values of x,} \\ \Rightarrow 50 =211.49 - 20.96 \cos h \left(0.03291765 x\right) \\ \Rightarrow 50-211.49=- 20.96 \cos h \left(0.03291765 x\right) \\ \Rightarrow -161.49 =- 20.96 \cos h \left(0.03291765 x\right) \\ \Rightarrow 161.49 = 20.96 \cos h \left(0.03291765 x\right) \mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ \text{Note:}- \cos h \left(u\right) =\frac{e^u+e^{-u}}{2}, \text{then} \\ \cos h \left(0.03291765 x\right) =\frac{e^{\left(0.03291765 x\right)}+e^{-\left(0.03291765 x\right)}}{2} \\ \text{Take} v=\left(0.03291765 x\right) then \\ \cos h \left(v\right) =\frac{e^v+e^{-v}}{2}=\frac{e^v+{\displaystyle \frac{1}{e^v}}}{2} \\ \text{Again take} e^v =t \text{ then} \\ \cos h \left(v\right) =\frac{e^v+e^{-v}}{2}=\frac{e^v+{\displaystyle \frac{1}{e^v}}}{2}=\frac{t+{\displaystyle \frac{1}{t}}}{2}=\frac{t^2+1}{2t} \end{gathered}$$

Step:-3

by (1), we get

$$\begin{gathered} 161.49 = 20.96 \cos h \left(0.03291765 x\right) \\ \Rightarrow 161.49 = 20.96 \left(\frac{t^2+1}{2t}\right) \\ \Rightarrow 161.49 = 10.48 \left(\frac{t^2+1}{t}\right) \\ \Rightarrow 15.4094=\frac{t^2+1}{t} \\ \Rightarrow 15.4094 t= t^2+1 \\ \Rightarrow t^2 -15.4094 t+1 =0 \end{gathered}$$

Step:-4

Applying quadratic formula, we get

$$\begin{gathered} t =\frac{15.4094 \pm \sqrt{{\left(15.4094\right)}^2 -4\times 1\times 1}}{2\times 1} \\ t =\frac{15.4094 \pm \sqrt{233.4481}}{2} \\ t=\frac{15.4094 \pm 15.2790}{2}=0.0652, 15.3442 \\ t=0.0652, 15.3442 \\ \text{but we assume} \\ e^v=t \\ \Rightarrow v=\ln \left(t\right) \\ \Rightarrow 0.03291765 x=\ln \left(t\right) \\ \Rightarrow x= \frac{\ln \left(t\right)}{0.03291765} \\ \text{when t}=0.0652 \text{then} \\ \Rightarrow x=-82.9432 \approx -82.94 \\ \Rightarrow x=-82.94 \\ \text{when t}=15.3442 \text{then} \\ \Rightarrow x=82.9566 \approx 82.96 \\ \Rightarrow x=82.96 \end{gathered}$$

Answer:-

Smaller x-value $\left(x, y\right)=\left(-82.94, 50\right)$

Larger  x-value $\left(x, y\right)=\left(82.96, 50\right)$